In a box there are $b$ black, $w$ white, and $r$ red balls. We choose $n$ balls from the box with replacement where $n\leq \min(b,w,r)$. Find the probability that we get at least 2 black balls and at least 2 white balls.
Here is my solution but I am not sure.
Let $B$ be the event that we get at least 2 black balls; $W$ be the event that we get at least 2 white balls. $$P(B\cap W)=1-P(B^c\cup W^c).$$ Therefore, we need to calculate $$P(B^c\cup W^c)=P(B^c)+P(W^c)-P(B^c\cap W^c).$$ Let $t=w+b+r$.
Note that $$P(B^c)=P(\mbox{exactly 0 black ball})+P(\mbox{exactly 1 black ball})=\binom{n}{0}(\frac{b}{t})^0(\frac{w+r}{t})^n+\binom{n}{1}(\frac{b}{t})^1(\frac{w+r}{t})^{n-1}.$$
Similarly, $$P(W^c)=P(\mbox{exactly 0 white ball})+P(\mbox{exactly 1 white ball})=\binom{n}{0}(\frac{w}{t})^0(\frac{b+r}{t})^n+\binom{n}{1}(\frac{w}{t})^1(\frac{b+r}{t})^{n-1}.$$ And
$$P(B^c\cap W^c)=P(\mbox{exactly 0 white and 0 black ball})+P(\mbox{exactly 1 white and 0 black ball})+P(\mbox{exactly 0 white and 1 black ball})+P(\mbox{exactly 1 white and 1 black ball}).$$ So,
$$P(B^c\cap W^c)=\binom{n}{n}(\frac{b+w}{t})^0(\frac{r}{t})^n+\binom{n}{1}(\frac{w}{t})^1(\frac{r}{t})^{n-1}+\binom{n}{1}(\frac{b}{t})^1(\frac{r}{t})^{n-1}+2\binom{n}{2}(\frac{w}{t})^1(\frac{b}{t})^1(\frac{r}{t})^{n-2}.$$
Under the assumption that $r>=2$, your answer is correct, but it would be good to simplify the expressions, and put it into standard multinomial distribution form for much greater transparency.
Since drawals are with replacement, probabilities of drawing black, white, red can be taken to be constant at $\frac{b}{t}=p,\; \frac{w}{t}=q,\;\frac{r}{t} = s$, and
$P(B^c\cap W^c) =$
$$\binom{n}{0,0,n}p^0q^0s^n + \binom{n}{0,1,n-1}p^0q^1s^{n-1} +\binom{n}{1,0,n-1}p^1q^0s^{n-1} + \binom{n}{1,1,n-2}p^1q^1s^{n-2}$$
and the final answer is its complement